How the smallest eigenvalue change in a structured matrix? Here is an n-by-n structured matrix,
$$
        \begin{bmatrix}
        1 & 1-c & 1-2c & ..... & 1-nc \\
        1-c & 1 & 1-c  & ..... & 1-(n-1)c\\
        \vdots & \vdots & \vdots & \vdots & \vdots\\
        1-nc & 1-(n-1)c & 1-(n-2)c & ..... & 1\\
        \end{bmatrix}
$$
It's a symmetric toeplitz matrix. c is a constant less than $\frac{1}{n}$ which means all entries are positive. I have known that this matrix is positive definite. 
What I want to do is to prove that the smallest eigenvalue decrease with c. I have simulated how the smallest eigenvalue change using MATLAB. Now I need to prove that using analysis. Could anyone give me any tip about that please? Thanks so much! 
 A: Let your matrix be $T $.
The characteristic polynomial of $T$ is of the form
$$ P_n(\lambda,c) = \lambda^{n+1} - (n+1) \lambda^n + \sum_{k=0}^{n-1} (a_{n,k} c^{n+1-k} - b_{n,k} c^{n-k}) \lambda^k $$ 
for some constants $a_{n,k}$ and $b_{n,k}$.  It may be possible to 
compute a closed form for these.
Differentiating $P_n(\lambda,c) = 0$ implicitly, we get 
$$ \dfrac{d\lambda}{dc} = - \dfrac{\partial P/\partial c}{\partial P/\partial \lambda}$$
as long as the denominator is nonzero, and this will have constant sign as long as both numerator and denominator are nonzero.  
For $c=0$, the eigenvalues are $n+1$ (with multiplicity $1$) and $0$ (with multiplicity $n$.  For $0 < c < 1/n$, $T$ is positive definite so the  eigenvalues are positive.  Thus the lowest $n$ eigenvalues will be increasing as a function of $c$ for small $c$. 
The problem (and I don't see immediately how to solve it) is to show that when $\lambda$ is the lowest eigenvalue, $\partial P/\partial c$ and $\partial P/\partial \lambda$ don't hit $0$ before $c= 1/n$.
Here's a plot in the case $n=5$.  Note how the first and second eigenvalues cross at approximately $c = 0.33$, after which the lowest eigenvalue is decreasing.  But before $c = 1/5$ the first $5$ eigenvalues are increasing
as functions of $c$.

